Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $r \neq 0$. $k = \dfrac{4r + 20}{2r^2 + 8r - 10} \times \dfrac{-4r^2 + 24r - 20}{2r + 2} $
Explanation: First factor out any common factors. $k = \dfrac{4(r + 5)}{2(r^2 + 4r - 5)} \times \dfrac{-4(r^2 - 6r + 5)}{2(r + 1)} $ Then factor the quadratic expressions. $k = \dfrac {4(r + 5)} {2(r - 1)(r + 5)} \times \dfrac {-4(r - 1)(r - 5)} {2(r + 1)} $ Then multiply the two numerators and multiply the two denominators. $k = \dfrac {4(r + 5) \times -4(r - 1)(r - 5) } { 2(r - 1)(r + 5) \times 2(r + 1)} $ $k = \dfrac {-16(r - 1)(r - 5)(r + 5)} {4(r - 1)(r + 5)(r + 1)} $ Notice that $(r - 1)$ and $(r + 5)$ appear in both the numerator and denominator so we can cancel them. $k = \dfrac {-16\cancel{(r - 1)}(r - 5)(r + 5)} {4\cancel{(r - 1)}(r + 5)(r + 1)} $ We are dividing by $r - 1$ , so $r - 1 \neq 0$ Therefore, $r \neq 1$ $k = \dfrac {-16\cancel{(r - 1)}(r - 5)\cancel{(r + 5)}} {4\cancel{(r - 1)}\cancel{(r + 5)}(r + 1)} $ We are dividing by $r + 5$ , so $r + 5 \neq 0$ Therefore, $r \neq -5$ $k = \dfrac {-16(r - 5)} {4(r + 1)} $ $ k = \dfrac{-4(r - 5)}{r + 1}; r \neq 1; r \neq -5 $